Problem: For the ellipse $16x^2 - 64x + y^2 + 4y + 4 = 0,$ find the distance between the foci.
Solution: Completing the square in $x$ and $y,$ we get
\[16(x - 2)^2 + (y + 2)^2 = 64.\]Then
\[\frac{(x - 2)^2}{4} + \frac{(y + 2)^2}{64} = 1.\]Thus, $a = 8$ and $b = 2,$ so $c = \sqrt{a^2 - b^2} = \sqrt{60} = 2 \sqrt{15}.$  Therefore, the distance between the foci is $2c = \boxed{4 \sqrt{15}}.$